A challenging integral

[Tolaso J Kos]

Prove that:

$$\int_{\pi}^{2\pi}\frac{\sin x}{e^x-\sin x-\cos x}\, dx=-\frac{\pi}{2}+\frac{1}{2}\ln \left ( e^\pi-1 \right )$$

[Grigorios Kostakos]

\begin{align*}
\displaystyle \int_{\pi}^{2\pi}{\frac{\sin x}{e^x-\sin x-\cos x} dx}&=\int_{\pi}^{2\pi}{\frac{1}{2}\Bigl({\frac{e^x-\cos x+\sin x}{e^x-\sin x-\cos x}-1}\Bigr) dx}\\
&=\frac{1}{2}\int_{\pi}^{2\pi}{\frac{e^x-\cos x+\sin x}{e^x-\sin x-\cos x}dx}-\frac{1}{2}\int_{\pi}^{2\pi}{dx}\\
&=\frac{1}{2}\int_{\pi}^{2\pi}{\frac{(e^x-\sin x-\cos x)'}{e^x-\sin x-\cos x}dx}-\frac{\pi}{2}\\
&=\frac{1}{2}\Bigl[{\log(e^x-\sin x-\cos x)}\Bigr]_{\pi}^{2\pi}-\frac{\pi}{2}\\
&=\frac{1}{2}\bigl({\log(e^{2\pi}-1)-\log(e^{\pi}+1)}\bigr)-\frac{\pi}{2}\\
&=\frac{1}{2}\log\bigl({\tfrac{e^{2\pi}-1}{e^{\pi}+1}}\bigr)-\frac{\pi}{2}\\
&=\frac{1}{2}\log({e^{\pi}-1})-\frac{\pi}{2}\,.
\end{align*}